---
title: "Zener current and power dissipation"
subject: "Fundamental of Electrical and Electronics Engineering"
chapter: 4
key: "semester_1/fundamental_of_electrical_and_electronics_engineering:zener-current-power"
source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/zener-current-power
site: BCT Engineering
---

# Zener current and power dissipation

Fundamental of Electrical and Electronics Engineering · Chapter 4: Diodes

- Asked in: 2081 Chaitra (new course), 4 marks
- Video: [Zener Diodes](https://www.youtube.com/watch?v=mmiHX_IzvDw) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/zener-current-power
- Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-4

## Question

A load resistance of 1 kΩ is connected across a Zener voltage regulator with Zener diode of 10 V. Find the Zener current and power dissipated across the Zener diode. [Figure: supply VS = 18 V in series with RS = 250 Ω feeding a 10 V Zener diode (VZ = 10 V, reverse biased) connected in parallel with load RL = 1 kΩ.]

## Short answer

Series current Is = (18−10)/250 = 32 mA, load current IL = 10 mA. Zener current Iz = 32−10 = 22 mA. Power dissipated Pz = 10 V × 22 mA = 220 mW.

## Answer

The Zener diode holds the load voltage at Vz = 10 V (it is in breakdown because the open-circuit load voltage 18 × 1000/1250 = 14.4 V is greater than 10 V).

```text
 Vs=18V o--[ Rs=250Ω ]--+--------+
                        |        |
                      Zener    RL=1kΩ
                      Vz=10V     |
        o---------------+--------+
```

#### Current through Rs

```text
Is = (Vs − Vz)/Rs = (18 − 10)/250 = 0.032 A = 32 mA
```

#### Load current

```text
IL = Vz/RL = 10/1000 = 0.010 A = 10 mA
```

#### Zener current (KCL)

```text
Iz = Is − IL = 32 − 10 = 22 mA
```

#### Power dissipated in the Zener

```text
Pz = Vz × Iz = 10 × 0.022 = 0.22 W = 220 mW
```

**Answer: Iz = 22 mA, Pz = 220 mW.**

(Power in Rs = 8 × 0.032 = 0.256 W; power in the load = 0.1 W.) The Zener should be rated above 220 mW, e.g. a 10 V, 0.5 W Zener.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the Zener in breakdown?
   A. The supply is 18 V
   B. Open-circuit load voltage 14.4 V exceeds 10 V
   C. RL is 1 kohm
   D. The Zener is forward biased
   Answer: B. 18 x 1000/1250 = 14.4 V > Vz, so it breaks down and holds 10 V. (Key point: Zener in breakdown)

2. What current flows through Rs?
   A. 18 mA
   B. 10 mA
   C. 32 mA
   D. 72 mA
   Answer: C. Is = (18 - 10)/250 = 32 mA. (Key point: Current through Rs)

3. Which gives Iz and Pz?
   A. Iz = 32 mA, Pz = 320 mW
   B. Iz = 10 mA, Pz = 100 mW
   C. Iz = 22 mA, Pz = 396 mW
   D. Iz = 22 mA, Pz = 220 mW
   Answer: D. Iz = Is - IL = 32 - 10 = 22 mA; Pz = 10 x 0.022 = 0.22 W. (Key point: Zener current and power)

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